REAL ANALYSIS I HOMEWORK 2

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1 REAL ANALYSIS I HOMEWORK 2 CİHAN BAHRAN The questions are from Stein and Shakarchi s text, Chapter Prove that the Cantor set C constructed in the text is totally disconnected and perfect. In other words, given two distinct points x, y C, there is a point z / C that lies in between x and y, and yet C has no isolated points. [Hint: If x, y C and x y > 1/3 k, then x and y belong to two different intervals in C k. Also, given any x C there is an end-point y k of some interval in C k that satisfies x y k and x y k 1/3 k.] Given two distinct points x, y C, since x y > 0 there exists k N such that x y > 1/3 k. Because C k is a union of closed intervals of length 1/3 k and x, y C k, there exists z in between x and y which does not lie in C k. Thus z / C. And given x C, for every k N, x lies in one of the 2 k intervals that make up C k. Thus there exists an end-point y k in C k such that x y k 1/3 k. We may assume y k x here (if x happens to be an end-point of an interval in C k itself, choose the other end-point of the interval to be y k ). But we know that the end-points survive the Cantor intersection, that is they lie in C. Hence [x 1/3 k, x + 1/3 k ] {x} intersects C for every k. Since x C was arbitrary, we conclude that C has no isolated points. 2. The Cantor set C can also be described in terms of ternary expansions. (a) Every number in [0, 1] has a ternary expansion x = a k 3 k, where a k = 0, 1 or 2. k=1 Note that this decomposition is not unique since, for example, 1/3 = k=2 2/3 k. Prove that x C if and only if x has a representation as above where every a k is either 0 or 2. The question states that there is a well-defined surjective map f : {0, 1, 2} N [0, 1] (a n ) a n 3 n. The question asks us to prove f({0, 2} N ) = C. In the last homework, I showed that f Ä {0, 2} Nä C (I also showed that f maps {0, 2} N injectively to C and from here deduced that C is uncountable). So it s enough to show that C f Ä {0, 2} Nä, or equivalently [0, 1] f Ä {0, 2} Nä [0, 1] C. So let x [0, 1] f Ä {0, 2} Nä. Since f is surjective there exists a sequence (a n ) with a n {0, 1, 2} such that x = a n 3 n. By assumption, the set {n N : a n = 1} must be nonempty. So pick the least element k from that set. Thus a k = 1 but 1

3 REAL ANALYSIS I HOMEWORK 2 3 And if we let 2 to be the sequence consisting entirely of 2 s and 1 to be the sequence of 1 s, we have thus (g ι)(2) = g(1) = 2 n = 1 = 2 3 n = f(2) F (1) = (g ι)(f 1 (1)) = (g ι)(2) = 1. For continuity, consider the discrete topology on {0, 1} and then endow the set of sequences {0, 1} N with the product topology. Do the same for {0, 2} N. Then being continuous coordinate-wise, ι is definitely continuous. We will show that f and g are also continuous. From here it follows that F is continuous, and since the domain {0, 2} N of f is compact (Tychonoff s theorem!) and the codomain C of f is Hausdorff, the bijective continuous map f is a homeomorphism; hence F is a composite of continuous maps. And Proposition 2 yields that f and g are continuous (g by taking d = 2 and f by taking d = 3 and restricting to {0, 2} N ). Lemma 1. Let d be a natural number greater than 1 and write D = {0,..., d 1}. Let (a n ) and (b n ) be two distinct sequences with elements in D. Let k = min{n N : a n b n }. Then b n d n a n d n < if and only if one of the three possibilities below holds: (1) a k < b k 1. (2) a k = b k 1 and b l > 0 for some l > k. (3) a k = b k 1 and a l < d 1 for some l > k. Proof. Omitted. This is a standard base-d expansion result. Proposition 2. Let d be a natural number greater than 1. Put the discrete topology on the set D = {0,..., d 1} and consider D N with the product topology. Then the surjective map is continuous. h : D N [0, 1] (a n ) a n d n Proof. It is enough to show that for every b [0, 1] the set h 1 ((, b) [0, 1]) = {(a n ) : a n d n < b} is open in D N. Pick (b n ) such that b n d n = b. If b n s eventually become 0, replace them by a sequence which is eventually d 1 which still is an expansion of b. Write U k = {b 1 } {b k 1 } {c D : c < b k 1} D D D V k,l = {b 1 } {b k 1 } {b k 1} D D D (D {d 1}) D D

4 REAL ANALYSIS I HOMEWORK 2 4 for every k, l N with l > k where the D {d 1} term in V k,l occurs at the l-th coordinate. Note that every U k and V k,l is open in D N. Now by Lemma 1 we get that {(a n ) : a n d n < b} = U k V k,l k N k N l>k which is open. (c) Prove that F : C [0, 1] is surjective. We know that f is bijective and g is surjective. ι is also clearly surjective by its definition. Thus F = g ι f 1 is also surjective. (d) One can also extend F to be a continuous function on [0, 1] as follows. Note that if (a, b) is an open interval of the complement of C, then F (a) = F (b). Hence we may define F to have constant value on that interval. A connected component of the complement of C is of the form ( n ) n a i 3 i + 3 n, a i 3 i n for some a 1,..., a n {0, 2}. Write r = Note that and as desired. n n r = a i 3 i i C i=n+1 n F (r) = (a i /2)2 i + 2 i i=n+1 n = (a i /2)2 i + 2 n n 1 = (a i /2)2 i + a n n 2 ( n 1 ) = F a i 3 i + (a n + 2)3 n ( n ) = F a i 3 i n = F (r + 3 n ) 5. Suppose E is a given set, and O n is the open set Show: O n = {x R d : d(x, E) < 1/n}. (a) If E is compact, then m(e) = lim n m(o n ). a i 3 i + 3 n so the interval is (r, r + 3 n ).

5 REAL ANALYSIS I HOMEWORK 2 5 Note that since E is closed, it is measurable so writing m(e) is OK. Also note that (O n ) is a decreasing family of open sets. Let O be an arbitrary open set containing E and write C = R d O. So E and C are disjoint, moreover since E is compact and C is closed they are distant by a previous exercise. Hence d(c, E) 1/n for some n. That is, every point in C is at least 1/n away from E hence C O n =. This implies that O n O. So we showed the set {O n : n N} is cofinal in {O R d : E O-open} with respect to the reverse containment. Thus by the monotonicity of m the set {m(o n ) : n N} is cofinal in {m(o) : E O-open} with respect to. Thus their infima coincide: lim O n = inf{m(o n ) : n N} = inf{m(o) : E O-open} = m(e). n (b) However, the conclusion in (a) may be false for E closed and unbounded; or E open and bounded. For the first part, take E = N as a subset of R, which is closed and unbounded. In this case, O n = (N 1/2n, N + 1/2n). N N Since the union is disjoint, by countable additivity of m we have whereas m(e) = 0, being a countable set. m(o n ) = N N 1/n = For the second part, let (q n ) be an enumeration of Q [0, 1] and take E = (q n 4 n, q n + 4 n ). E is certainly open and bounded. Now by countable subadditivity we have m(e) 2 4 n = /4 = /4 = 2/3. However, E is dense in [0, 1]. So given x [0, 1], for every n N there exists q E such that d(x, q) < 1/n so x O n. Thus [0, 1] O n for every n and hence for every n. Therefore 1 = m([0, 1]) m(o n ) lim m(o n) 1. n 11. Let A be the subset of [0, 1] which consists of all numbers which do not have the digit 4 appearing in their decimal expansion. Find m(a). This corresponds to dividing [0, 1] into 10 equal pieces, labeling them from 0 to 9 and taking away the piece with label 4 (since the numbers between 0.4 and 0.5 should be deleted). Then this process is applied to the 9 remaining pieces. So at the first step

6 REAL ANALYSIS I HOMEWORK 2 6 we take away an interval of length 1/10, at the second step we take away 9 intervals of length 1/100 and so on. So the complement of A has measure 9 n = 1 n 10 Ç å 9 n 1 = = 1 10 hence A has measure The following deals with G δ and F σ sets. (a) Show that a closed set is a G δ and an open set an F σ. [Hint: If F is closed, consider O n = {x : d(x, F ) < 1/n}.] Let s use the notation of the hint. Since F is closed, d(x, F ) = 0 if and only if x F therefore F = O n. Let G be an open set. Then G c is closed and hence is a G δ set. Therefore G is an F σ set by de Morgan. (b) Give an example of an F σ that is not a G δ. [Hint: This is more difficult; let F be a denumerable set that is dense.] We can definitely choose an F as in the hint for any R d because R d is a separable topological space (points with rational coordinates form a dense set). Since F is a countable union of singletons, it is an F σ. To show that it is not G δ we will refer to the Baire category theorem. Definition 3. A topological space X is called a Baire space if for any countable collection {A n } of closed sets of X each of which has empty interior in X, their union An also has empty interior in X. Theorem 4. (Baire category theorem) Complete metric spaces are Baire spaces. Now suppose F is G δ to get a contradiction. Then F = U n where U n s are a countable collection of open sets. Since F is dense, each U n is dense. Write C n = R d U n. Then since Int(C n ) is disjoint from the dense set U n, we have Int(C n ) =. So R d F = C n is a countable union of closed sets with empty interiors. But F is also a countable union of closed sets with empty interiors, namely singletons. Thus R d = (R d F ) F is a countable union of closed sets with empty interiors. But by Baire category theorem R d is a Baire space hence R d has empty interior, nonsense. Therefore F is not G δ. (c) Give an example of a Borel set which is not a G δ nor an F σ. Lemma 5. Let X be any topological space and A, B X. (1) If A and B are F σ sets, so are A B and A B.

7 REAL ANALYSIS I HOMEWORK 2 7 (2) If A and B are G δ sets, so are A B and A B. Proof. (2) follows from (1) by taking complements. To prove (1), write A = F n and B = L n where F n, L n are closed. Then A B = (F n L n ), A B = (F n L m ) are F σ sets. n,m N Let F = [0, ) Q and G = (, 0] (R Q) in R. Note that closed sets in R are trivially F σ and also G δ by part (a). By Lemma 5, F is an F σ set and G is a G δ set. Let E = F G. Being the union of two Borel sets, E is certainly Borel. Suppose that E is G δ. Then by Lemma 5 the set E [0, ) = F is G δ. But then the set F = (, 0] Q is also G δ because F is the image of F under the homeomorphism x x of R. Thus F F = Q is G δ. This contradicts what we ve shown in part (b). Thus E is not G δ. Now suppose that E is F σ. Then E (, 0] = G is F σ. Similar to above, from here it follows that R Q is F σ which then implies that Q is G δ, again a contradiction. Thus E is neither G δ nor F σ. 16. The Borel-Cantelli lemma. Suppose {E k } k=1 is a countable family of measurable subsets of R d such that m(e k ) <. Let k=1 = lim sup(e k ). k (a) Show that E is measurable. [Hint: Write E = n=1 k n E k ]. E = {x R d : x E k, for infinitely many k} Let s verify the hint. Assume x E k for infinitely many k. This implies that for every n N, there exists k n such that x E k. That is, for every n N, x k n E k and hence x k n E k. Conversely, assume that x k n E k. Since x k 1 E k, there exists k 1 1 such that x E k1. And since x k 2 E k, there exists k 2 > max{2, k 1 } such that x E k2. Going on like this, we can construct a sequence k 1 < k 2 < k 3 such that x E kn for every n. Thus x lies in infinitely many E k. Having proved the claim, we are done since the collection of measurable sets is closed under countable unions and countable intersections. (b) Prove m(e) = 0. Since k=1 m(e k ) <, for every ε > 0 there exists n N such that Ñ é ε > m(e k ) m m(e). k n E k k n

8 REAL ANALYSIS I HOMEWORK 2 8 Thus m(e) = Let {f n } be a sequence of measurable functions on [0, 1] with f n (x) < for a.e. x. Show that there exists a sequence c n of positive real numbers such that f n (x) c n 0 a.e. x [Hint: Pick c n such that m({x : f n (x)/c n > 1/n}) < 2 n, and apply the Borel- Cantelli lemma.] Fix n. Since f n < a.e., for every n there exists d n > 0 such that m({x : f n (x) > d n }) < 2 n. Now let c n = nd n. Clearly (c n ) is a sequence of positive real numbers. If we write E n = {x : f n (x) > d n } = x : f n(x) > 1/n, c n we have m(e n ) < 2 n = 1 <. So by the Borel-Cantelli lemma, the E := lim sup n (E n ) is a null set. Now by definition, every x E c lies in only finitely many E n s. That is, given x E c, there exists N N such that for every n N we have Therefore f n(x) c n 0 for every x E c. f n (x) c n 1/n. 29. Suppose E is a measurable subset of R with m(e) > 0. Prove that the difference set of E, which is defined by {x y : x, y E} contains an open interval centered at the origin. It is enough to prove the claim for a measurable subset of E with positive measure, so we do some reductions by finding some nice subsets of E like this and replacing E with these subsets. First, since the collection {E (n, n + 1] : n Z} of disjoint measurable sets cover E, by countable additivity there exists n N such that m(e (n, n + 1]) > 0. So we may assume that E has finite measure. Second, since 0 < m(e) <, by an exercise in the previous homework there exists a compact set K contained in E such that (choosing ε = m(e)/2) m(e K) m(e)/2. Then by addivity of the measure we get m(k) m(e)/2 > 0. So we may assume that E is compact.

10 REAL ANALYSIS I HOMEWORK 2 10 that x + a + c E. Therefore a + c (E, F ). Since this is true whenever 0 < c < δ, the interval (a δ, a + δ) is contained in (E, F ). 33. Let N denote the non-measurable set constructed in the text. Recall from the exercise above that measurable subsets of N have measure zero. Show that the set N c = I N satisfies m (N c ) = 1, and conclude that if E 1 = N and E 2 = N c, then although E 1 and E 2 are disjoint. m (E 1 ) + m (E 2 ) m (E 1 E 2 ) [Hint: To prove that m (N c ) = 1, argue by contradiction and pick a measurable set U such that U I, N c U and m (U) < 1 ε.] Suppose m (N c ) < 1, so there exists ε > 0 such that m (N c ) < 1 ε. Since m (N c ) = inf{m(u) : N c U-open} there exists an open, hence measurable set U containing N c such that m(u) < 1 ε. Note that U I is also a measurable containing N c with m(u I) < 1 ε, so we may assume U I. Since N c = I N U I, we have I U N. But I U is a measurable set so by additivity of the measure we have m(i U) = m(i) m(u) = 1 m(u) > ε. So I U is a measurable subset of N with positive measure; a contradiction. Therefore m (N c ) 1 but on the other hand m (N c ) m (I) = 1 hence m (N c ) = 1. We know that sets with outer measure zero are measurable, so m (N ) > 0. Thus m (N ) + m (N c ) > 1 = m (I) = m (N N c ). 34. Let C 1 and C 2 be any two Cantor sets (constructed in Exercise 3). Show that there exists a function F : [0, 1] [0, 1] with the following properties: (i) F is continuous and bijective. (ii) F is monotonically increasing. (iii) F maps C 1 surjectively onto C 2. [Hint: Copy the construction of the standard Cantor-Lebesgue function.] Let C be a Cantor set of constant dissection as in Exercise 3. By construction, C is the intersection of a family {C n } of closed sets where each C n is a disjoint union of 2 n closed intervals. So we can label these 2 n intervals from left to right by bit strings of length n, that is, words of length n consisting of 0 s and 1 s. So for example C 1 = I 0 I 1 where I 0 is the interval on the left hand side in C 1 and I 1 is the one on the right. Keeping the labeling in a lexicographic order, we have C 2 = I 00 I 01 I 10 I 11 and in general C n is the union of I b s where b s vary over length n bit strings. Note that I b I c if and only if c can be truncated from the right to obtain b. For example I 0 I 01 I 010 I In general given an infinite sequence a = (a n ) of 0 s and 1 s, if we write a n for its n-truncation (a 1,..., a n ) there is a decreasing sequence I a 1 I a 2 I a 3

11 By compactness, the intersection REAL ANALYSIS I HOMEWORK 2 11 I a n which lies in C, is nonempty. Yet the diameter of the intersection is zero, hence it must be a singleton. Therefore every infinite sequence a of 0 s and 1 s uniquely determines a point in C. So we get a map f : {0, 1} N C which is surjective since points in C by definition survives the intersection of C n s hence lie in infinitely many (hence in an infinite decreasing chain of) I b s. If two infinite sequences are distinct, they have different truncations so as the intervals get finer, the two points these sequences determine will fall into different intervals. Hence f is a bijection. Two points in C lie in the same I b where b is a finite bit string if and only if their inverse images under f both start with b. It follows from this observation (as we did for the middle thirds Cantor set in Exercise 2) that f is continuous. And since f goes from a compact space to a Hausdorff space, f is a homeomorphism. Also observe that if we order {0, 1} N by lexicographic ordering, then f preserves the order. Because if a sequence beats another sequence lexicographically, then at some point it will lie to the right side of a dissection while the other lies on the left side. So if C 1 and C 2 are two Cantor sets, we have order preserving homeomorphisms f 1 : {0, 1} N C 1 and f 2 : {0, 1} N C 2 ; thus f 2 f1 1 gives an order preserving homeomorphism from C 1 to C Give an example of a measurable function and a continuous function Φ so that f Φ is non-measurable. [Hint: Let Φ : C 1 C 2 as in Exercise 34, with m(c 1 ) > 0 and m(c 2 ) = 0. Let N C 1 be non-measurable, and take f = χ Φ(N).] Use the construction in the hint to show that there exists a Lebesgue measurable set that is not a Borel set. Let s do as the hint commands. We know such N exists by Exercise 32(b). Since Φ(N) C 2 and m(c 2 ) = 0, we have m (Φ(N)) = 0 and so Φ(N) is a measurable set. Therefore f is a measurable function. However, (f Φ) 1 (1) = Φ 1 (f 1 ({1})) = Φ 1 (Φ(N)) = N is not measurable, hence f Φ is not a measurable function. Also the measurable set Φ(N) cannot be Borel because the inverse images of Borel sets under continuous functions are Borel, but although Φ is continuous, Φ 1 (Φ(N)) = N is not even measurable.

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